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I think the problem statement is clear, and is independent of how the TV show actually operated. The problem posed is that you have three closed doors, behind

I think the problem statement is clear, and is independent of how the TV show actually operated. The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car). One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is either behind the door you picked or behind the other closed door. In this setup, the question is: if offered the chance to switch your selection from the door you initially picked to the other closed door, should you switch? Does it make a difference? Mariyln Vos Savant said switch. The peanut gallery (some statisticians who should have known better) said it makes no difference - with two doors left you've got a 50/50 chance of winning the car whether you switch or not. First off, you can run multiple simulations of the game (with random positions of the goats and car), and prove for yourself that always switching is a better strategy, so it's sad that anyone would insist that it's not, even if that seems logical. The simplest way to explain why switching is the correct strategy is this: When you make your initial selection, you have: a) A 1/3 chance of selecting the car, and b) A 2/3 chance of selecting one of the two goats So, 1/3 of the time you selected the car and will therefore lose if you switch, and 2/3 of the time you selected a goat and will therefore win if you switch. Switching will therefore be a winning strategy 2/3 of the time.

petesergeant 3y ago
> The simplest way to explain why switching is the correct strategy In my experience the most _intuitive_ explanation is to simply ramp it up to 100 doors, with Monty opening 98 of them, to make it clear that switching offers you the benefit of all unopened doors.
- HarHarVeryFunny 3y ago
  Maybe different explanations resonate with different people. Another explanation: After initial selection there is 1/3 chance car is behind your door, and 2/3 chance it is behind one of the other two doors. When one of the other doors is opened revealing a goat, we NOW know (new information to be taken advantage of!) the 2/3 chance represented by those two other doors lies with the unopened one. So, your choice is stick with original choice which has 1/3 chance of being correct, or switching to the other door which you now know has a 2/3 chance of being correct! Obviously you switch. The big lie to "with two doors left it's 50/50" is that opening doors makes no difference. You're original choice had 1/3 chance of being correct, and you can't go back in time and change that.
  - lynguist 3y ago
    “People” seem to think whenever you have a number of options, each option is equally likely. I get into the car today - either I die in a crash or I don’t. Hmm, I didn’t die, I survived the 50:50. Their model doesn’t allow accounting for varying probabilities for different outcomes.
heleninboodler 3y ago
It's a lot easier to see the 2/3 nature if you imagine that instead of revealing the goat and then asking if you want to switch, Monty said "you've chosen this door that has a 1/3 chance. I'm offering you to switch to choosing both of these two remaining doors, at least one of which definitely has a goat by definition." This is functionally the same offer. It's 2/3 to switch because you're effectively choosing two doors and having Monty automatically grant you the car if it's behind one of them.
- listenallyall 3y ago
  The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3. Imagine instead: After the player picks a door, the host states out loud, "we started with 2 zonks, you picked one door, so there must be at least one unchosen zonk... and I'm going to show you one behind door #X (door opens with zonk)... now, would you like to switch?" Wouldn't make for good TV but it describes what is actually happening. Knowing that the host will always show an unchosen zonk is the key to realizing that opening that door has no impact on the player's odds (1/3 originally, 1/3 after the reveal, thus switching is 2/3).
  - heleninboodler 3y ago
    > The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3 Yes, exactly. Which means when you switch, the host is effectively allowing you to choose both the other doors, which gives you 2/3 odds.
listenallyall 3y ago
It absolutely is not "independent of how the show operated." The host knowing in advance where the car is, (thereby ensuring the revealed door is always a zonk) is essential. If the host just picked a random door every day, and sometimes did reveal the car prematurely, then it would in fact be a 50-50 decision to switch.
- mathgeek 3y ago
  The math here, for those wondering, is that “informed host has a 0 chance to pick the car” becomes “ignorant host has a 1/3 chance to pick the car”. Contestant’s original pick is still 1/3 and the unpicked door goes from 2/3 (informed) to 1/3 (ignorant).
- HarHarVeryFunny 3y ago
  That doesn't change anything. Obviously if the host reveals the car then you've lost, but we're not talking about such a case. Regardless of whether that is happening some of the time, there is always a 2/3 chance your initial pick was wrong and you'd be better off switching.
  - listenallyall 3y ago
    If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates." Another show, Deal or No Deal, often came down to 2 suitcases (from an initial 26), one of which was worth a six-figure amount and the other a tiny amount. Something like $200,000 vs $500. The logic of the Monty Hall problem would suggest that since the player had 1/26 chance of initially choosing $200,000, they should switch suitcases at the end. But because all 24 previous reveals were random, there is actually no advantage to doing so.
    - HarHarVeryFunny 3y ago
      > If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of there being a car anywhere, when we know for sure (3/3 chance) that it is somewhere! The reality is that the contestant's original pick has a 1/3 chance of being right, and the other other two doors have a combined 2/3 chance of the car being behind one of them. Once one of those doors has been opened and the car shown not to be there, then this now means this combined 2/3 chance lies with the other door (which you should therefore switch to). Note that the probabilities now add up as expected: 1/3 chance original pick + 2/3 remaining choice = 3/3.

petesergeant 3y ago

> The simplest way to explain why switching is the correct strategy In my experience the most _intuitive_ explanation is to simply ramp it up to 100 doors, with Monty opening 98 of them, to make it clear that switching offers you the benefit of all unopened doors.

HarHarVeryFunny 3y ago
Maybe different explanations resonate with different people. Another explanation: After initial selection there is 1/3 chance car is behind your door, and 2/3 chance it is behind one of the other two doors. When one of the other doors is opened revealing a goat, we NOW know (new information to be taken advantage of!) the 2/3 chance represented by those two other doors lies with the unopened one. So, your choice is stick with original choice which has 1/3 chance of being correct, or switching to the other door which you now know has a 2/3 chance of being correct! Obviously you switch. The big lie to "with two doors left it's 50/50" is that opening doors makes no difference. You're original choice had 1/3 chance of being correct, and you can't go back in time and change that.
- lynguist 3y ago
  “People” seem to think whenever you have a number of options, each option is equally likely. I get into the car today - either I die in a crash or I don’t. Hmm, I didn’t die, I survived the 50:50. Their model doesn’t allow accounting for varying probabilities for different outcomes.

heleninboodler 3y ago

It's a lot easier to see the 2/3 nature if you imagine that instead of revealing the goat and then asking if you want to switch, Monty said "you've chosen this door that has a 1/3 chance. I'm offering you to switch to choosing both of these two remaining doors, at least one of which definitely has a goat by definition." This is functionally the same offer. It's 2/3 to switch because you're effectively choosing two doors and having Monty automatically grant you the car if it's behind one of them.

listenallyall 3y ago
The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3. Imagine instead: After the player picks a door, the host states out loud, "we started with 2 zonks, you picked one door, so there must be at least one unchosen zonk... and I'm going to show you one behind door #X (door opens with zonk)... now, would you like to switch?" Wouldn't make for good TV but it describes what is actually happening. Knowing that the host will always show an unchosen zonk is the key to realizing that opening that door has no impact on the player's odds (1/3 originally, 1/3 after the reveal, thus switching is 2/3).
- heleninboodler 3y ago
  > The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3 Yes, exactly. Which means when you switch, the host is effectively allowing you to choose both the other doors, which gives you 2/3 odds.

listenallyall 3y ago

It absolutely is not "independent of how the show operated." The host knowing in advance where the car is, (thereby ensuring the revealed door is always a zonk) is essential. If the host just picked a random door every day, and sometimes did reveal the car prematurely, then it would in fact be a 50-50 decision to switch.

mathgeek 3y ago
The math here, for those wondering, is that “informed host has a 0 chance to pick the car” becomes “ignorant host has a 1/3 chance to pick the car”. Contestant’s original pick is still 1/3 and the unpicked door goes from 2/3 (informed) to 1/3 (ignorant).
HarHarVeryFunny 3y ago
That doesn't change anything. Obviously if the host reveals the car then you've lost, but we're not talking about such a case. Regardless of whether that is happening some of the time, there is always a 2/3 chance your initial pick was wrong and you'd be better off switching.
- listenallyall 3y ago
  If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates." Another show, Deal or No Deal, often came down to 2 suitcases (from an initial 26), one of which was worth a six-figure amount and the other a tiny amount. Something like $200,000 vs $500. The logic of the Monty Hall problem would suggest that since the player had 1/26 chance of initially choosing $200,000, they should switch suitcases at the end. But because all 24 previous reveals were random, there is actually no advantage to doing so.
  - HarHarVeryFunny 3y ago
    > If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of there being a car anywhere, when we know for sure (3/3 chance) that it is somewhere! The reality is that the contestant's original pick has a 1/3 chance of being right, and the other other two doors have a combined 2/3 chance of the car being behind one of them. Once one of those doors has been opened and the car shown not to be there, then this now means this combined 2/3 chance lies with the other door (which you should therefore switch to). Note that the probabilities now add up as expected: 1/3 chance original pick + 2/3 remaining choice = 3/3.